Consider: f(x) = tan⁻¹(√((1+sin x)/(1-sin x))), x in (0, π/2). A normal to y = f(x) at x = π/6 also

Q: Wait, why do we reject the negative root when simplifying the square root?

Great observation! When we simplify $\sqrt{(\cos(x/2) + \sin(x/2))^2 / (\cos(x/2) - \sin(x/2))^2}$, we technically get $|(\cos(x/2) + \sin(x/2)) / (\cos(x/2) - \sin(x/2))|$. However, we are given the interval $x \in (0, \pi/2)$, which implies $x/2 \in (0, \pi/4)$. In this range, $\cos(x/2) > \sin(x/2)$, ensuring the denominator is positive. Thus, the expression inside the absolute value is already positive, and we can safely remove the absolute value bars.

Q: Why can't we just use the chain rule on the original function?

You absolutely could, but it would be a calculation nightmare! The chain rule would involve differentiating the square root, then the quotient, and then the inverse tangent. By simplifying the function first, we transform a complex calculus problem into a simple linear derivative. In JEE, always look for the simplification path first—it saves time and prevents errors.

Q: How do I find the slope of the normal if the tangent slope is $1/2$?

Remember the geometric property of perpendicular lines: if two lines are perpendicular, the product of their slopes is $-1$. If the slope of the tangent is $m_t = 1/2$, then the slope of the normal $m_n$ must satisfy $m_t \cdot m_n = -1$. Therefore, $m_n = -1 / (1/2) = -2$. It is a simple but crucial step!

Question

Consider: $f(x) = 	an^{-1} \left( \sqrt{\frac{1+\sin x}{1-\sin x}} ight), x \in \left( 0, \frac{\pi}{2} ight)$. A normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point

Options:

(A) (\pi/6, 0)

(B) (\pi/4, 0)

(C) (0, 0)

(D) (0, 2\pi/3)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are staring at the function $f(x) = 	an^{-1} \left( \sqrt{\frac{1+\sin x}{1-\sin x}} ight)$ for $x \in (0, \frac{\pi}{2})$. Your first instinct might be to reach for the chain rule, but let's pause.

In the world of JEE Advanced, brute force is rarely the intended path. Instead, we look for the hidden geometric elegance. The expression inside the square root is a classic trap.

We know that $1 = \cos^2(x/2) + \sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$. By substituting these, we can rewrite $1 \pm \sin x$ as $(\cos(x/2) \pm \sin(x/2))^2$.

This transforms our function into:

$$f(x) = 	an^{-1} \left( \sqrt{\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))^2}} ight)$$

### The Beauty of the Interval

Now, we must be careful. When we take the square root, we must consider the interval. Since $x \in (0, \frac{\pi}{2})$, we know $x/2 \in (0, \frac{\pi}{4})$.

In this region, $\cos(x/2) > \sin(x/2)$, which means our denominator is positive. The square root simplifies cleanly to:

$$\frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}$$

To make this look like a tangent, we divide the numerator and denominator by $\cos(x/2)$, yielding:

$$\frac{1 + 	an(x/2)}{1 - 	an(x/2)}$$

This is the standard expansion for $	an(\frac{\pi}{4} + \frac{x}{2})$. Thus, our function simplifies to $f(x) = 	an^{-1}(	an(\frac{\pi}{4} + \frac{x}{2}))$.

Since our angle is within the principal domain, the function collapses into the beautiful, simple linear form:

$$f(x) = \frac{\pi}{4} + \frac{x}{2}$$

### The Calculus and the Geometry

Now, the calculus becomes trivial. The derivative is:

$$f'(x) = \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} ight) = \frac{1}{2}$$

This is the slope of our tangent, $m_t = 1/2$. The normal, being perpendicular, has a slope $m_n = -2$.

To find the equation of the normal, we need a point. At $x = \frac{\pi}{6}$:

$$y = f\left(\frac{\pi}{6}ight) = \frac{\pi}{4} + \frac{\pi/6}{2} = \frac{\pi}{4} + \frac{\pi}{12} = \frac{\pi}{3}$$

Our point is $(\frac{\pi}{6}, \frac{\pi}{3})$. Using the point-slope form $y - y_1 = m_n(x - x_1)$, we get:

$$y - \frac{\pi}{3} = -2\left(x - \frac{\pi}{6}ight)$$

This simplifies to the final equation:

**$y = -2x + \frac{2\pi}{3}$**

Testing our result, we see that when $x=0$, $y = \frac{2\pi}{3}$. The normal passes through $(0, \frac{2\pi}{3})$. We have successfully navigated the trap and found the solution with elegance.

Animated Solution for Mathematics - Differentiation: Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

Select Answer:

Visualized Solution

Visualizing the Function

Trigonometric Identity

Substitution in $f (x)$

Simplifying the Square Root

Converting to Tan Form

Final Form of $f (x)$

Finding the Slope of Tangent

Finding the Slope of Normal

Point of Tangency

Equation of the Normal

Final Verification

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Beauty of the Interval

The Calculus and the Geometry

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider: f(x)=tan−1(1−sinx1+sinx​​),x∈(0,2π​). A normal to y=f(x) at x=6π​ also passes through the point